A Carnot engine converts one fifth of the heat input into work. If the sink temperature is reduced by 80^o C, the efficiency gets doubled. Find the source and the sink temperature. Also give the consequences of the Carnot cycle.

Maximum efficiency of an engine working between temperatures T₂ and T₁ is given by the fraction of the heat absorbed by an engine which can be converted into work is known as efficiency of the heat engine.

Mathematically,
Efficiency, η = (T₂ – T₁) / T₂ = 1/5
5 T₂ -5 T₁ = T₂
Therefore, T₂ = 1.25 T₁ ……………………….. (1)
Where T₂ is the source temperature and T₁ is the sink temperature.

If the sink temperature (T₁) is reduced by 80^o C, the efficiency gets doubled i.e.
η= T₂ – (T₁ – 80) / T₂ = 2 X 1/5 = 2/5
5 T₂ – 5 T₁ + 400 = 2 T₂
5 T₂ – 2 T₂ – 5 T₁ + 400 = 0
3 T₂ – 5 T₁ + 400 = 0
Substituting the value of T₂ from equation (1), we will get,
3 (1.25 T₁) – 5 T₁ + 400 = 0
3.75 T₁ – 5 T₁ + 372 = 0
400 = 1.25 T₁
T₁ = 320 K
Therefore, T₂ = 1.25 X 320 = 400 K

Hence, temperature of the source and the sink is 400 K and 320 K respectively.

Consequences or impartibility of the Carnot cycle:
(1) It is impossible to perform a frictionless process.

(2) It is impossible to transfer the heat without finite temperature difference.

(3) During isothermal process the piston must move very slowly so that the temperature remains constant and during adiabatic process it must move very fast so that the heat transfer is negligible due to short span of time. The variation speed during same cycle is not possible.